Dokaži, da: -cot ^ -1 (theta) = cos ^ -1 (theta) / +1 + (theta) ²?

Let #cot ^ (- 1) theta = A # potem

# rarrcotA = theta #

# rarrtanA = 1 / theta #

# rarrcosA = 1 / secA = 1 / sqrt (1 + tan ^ 2A) = 1 / sqrt (1+ (1 / theta) ^ 2) #

# rarrcosA = 1 / sqrt ((1 + theta ^ 2) / theta ^ 2) = theta / sqrt (1 + theta ^ 2) #

# rarrA = cos ^ (- 1) (theta / (sqrt (1 + theta ^ 2))) = cot ^ (- 1) (theta) #

#rarrthereforecot ^ (- 1) (theta) = cos ^ (- 1) (theta / (sqrt (1 + theta ^ 2))) #

Pokažite, da cos²π / 10 + cos²4π / 10 + cos² 6π / 10 + cos²9π / 10 = 2. Malo sem zmeden, če naredim Cos²4π / 10 = cos² (π-6π / 10) & cos²9π / 10 = cos² (π-π / 10), bo postal negativen kot cos (180 ° - theta) = - costheta v drugi kvadrant. Kako naj dokazujem vprašanje?

Glej spodaj. LHS = cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10) + cos ^ 2 ((6pi) / 10) + cos ^ 2 ((9pi) / 10) = cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10) + cos ^ 2 (pi- (4pi) / 10) + cos ^ 2 (pi- (pi) / 10) = cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10) + cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10) = 2 * [cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10)] = 2 * [cos ^ 2 (pi / 2- (4pi) / 10) + cos ^ 2 ((4pi) / 10)] = 2 * [sin ^ 2 ((4pi) / 10) + cos ^ 2 ((4pi) / 10)] = 2 * 1 = 2 = RHS